SELINA Solution Class 9 Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram , Rectangle , Rhombus, Square and Trapezium) Exercise 14C

Question 1

E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.

Sol:

Let us draw a parallelogram ABCD Where F is the midpoint Of side DC and E is the mid-point of side AB of a parallelogram  ABCD.

To prove:  AEFD is a parallelogram

Proof: 
In parallelogram ABCD
AB || DC
BC || AD
AB = DC
${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$DC
AE = DF
Also AD || EF
Therefore, AEFD is a parallelogram.

Question 2

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

Sol:

Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram  ABCD at angle B and D

To Prove:  ABCD is a rhombus

Proof: Let us draw a parallelogram  ABCD where the diagonal BD bisects the parallelogram at an angle B and D.

Construction: Let us join AC as a diagonal of the parallelogram ABCD

Since ABCD  is a parallelogram
Therefore
AB = DC
AD =BC
Diagonal  BD bisects angle B and D
So ∠COD = ∠DOA
Again AC also bisects at A and C   

Therefore ∠AOB =∠ BOC
Thus ABCD is a rhombus.
Hence proved

Question 3

The alongside figure shows a parallelogram ABCD in which AE = EF = FC.
Prove that:
(i) DE is parallel to FB
(ii) DE = FB
(iii) DEBF is a parallelogram.

Sol:

Construction : 
Join DF and EB
Join diagonal BD

Since diagonals of a parallelogram bisect each other.
∴  OA = OC and OB = OD
Also, AE = EF = FC

Now, OA = OC and AE = FC
⇒  OA - AE = OC - FC
⇒  OE = OF

Thus, in quadrilatreal DEFB, bisect each other.
OB = OD and OE = OF
⇒  Diagonals of a quadrilateral DEFB bisect each other.
⇒ DEFB is a parallelogram.
⇒  DE is parallel to FB
⇒  DE = FB                      .....( Opposite sides are equal )

Question 4

In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that : 

(i) AQ = BP
(ii) PQ = CD.
(iii) ABPQ is a parallelogram.

Sol:

Let us join PQ. 
Consider the ΔAOQ and ΔBOP
∠AOQ = ∠BOP                  ..[ opposite angles ]
∠OAQ = ∠BPO                 ...[ alternate angles ]
⇒ ΔAOQ ≅ ΔBOP            ...[ AA test ]
Hence AQ = BP

Consider the ΔQOP and ΔAOB
∠AOB = ∠QOP                 ...[ opposite angles ]
∠OAB = ∠APQ                 ...[ alternate angles ]
⇒ ΔQOP ≅ ΔAOB              ...[ AA test ]
Hence PQ = AB = CD

Consider the quadrilateral QPCD
DQ = CP and DQ || CP ||       ...[ Since AAD = BC and AD || BC ]
Also QP = DC and AB || QP || DC

Hence Quadrilateral QPCD is a Parallelogram.

Question 5

In the given figure, ABCD is a parallelogram. Prove that: AB = 2 BC.

Sol:

Given ABCD is a parallelogram
To prove: AB = 2BC

Proof:  ABCD is a parallelogram
A + D + B + C = 180°

From the AEB we have
⇒ ${\displaystyle \frac{\text{∠A}}{2}+\frac{\text{∠B}}{2}}$ + E = 180°
⇒ ∠A - ${\displaystyle \frac{\text{∠A}}{2}}$ + ∠D + ∠E1 = 180° ...[ taking E1 as new angle ]
⇒ ∠A + ∠D + ∠E1 = 180° + ${\displaystyle \frac{\text{∠A}}{2}}$ 
⇒ ∠E1 = ${\displaystyle \frac{\text{∠A}}{2}}$  ...[ Since ∠A + ∠D = 180°

Again,
similarly ,
∠E1 = ${\displaystyle \frac{\text{∠B}}{2}}$ 
Now
AB = DE + EC
= AD + BC
= 2BC                                ...[ since AD = BC]
Hence proved.

Question 6

Prove that the bisectors of opposite angles of a parallelogram are parallel.

Sol:

Given ABCD is a parallelogram. The bisectors of ∠ADC and ∠BCD meet at E. The bisectors of ∠ABC and ∠BCD meet at F

From the parallelogram ABCD we have

∠ADC + ∠BCD = 180° ...[ sum of adjacent angles of a parallalogram ]
⇒  ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°
⇒ ∠EDC + ∠EDC = 90°

In triangle ECD sum of angles = 180°
⇒ ∠EDC + ∠ECD + ∠CED = 180°
= ∠CED = 90°

Similarly taking triangle BCF it can prove that ∠BFC = 90°
Now since
∠BFC = ∠CED = 90°

Therefore the lines DE and BF are parallel
Hence proved

Question 7

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Sol:

Given:
ABCD is a parallelogram
AE bisects ∠BAD
BF bisects ∠ABC
CG bisects ∠BAD
DH bisects ∠ADC

To prove: LKJI is a rectangle

Proof :
∠BAD + ∠ABC = 180° ...[ adjacent angles of a parallelogram are supplementary ] 
∠BAJ = ${\displaystyle \frac{1}{2}}$ ∠BAD ...[AE bisects  BAD ]
∠ABJ = ${\displaystyle \frac{1}{2}}$ ∠ABC ... [DH  bisect ABC ]
∠BAJ + ∠ABJ = 90° ...[ halves of supplementary angles are complementary ]

ΔABJ is a right triangle because its acute interior angles are complementary.
Similarly
∠DLC = 90°
∠AID = 90°

Then ∠JIL = 90° because ∠AID and ∠JIL are vertical angles

since 3 angles of a quadrilateral, LKJI are right angles, si is the 4^th one and so is LKJI a rectangle, since its interior angles are all right angles
Hence proved.

Question 8

Prove that the bisectors of the interior angles of a rectangle form a square.

Sol:

Given: A parallelogram ABCD in which AR, BR, CP, DP 
Are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.

To prove: PQRS is a rectangle

Proof :
∠DCB + ∠ ABC =180° ...[ co - interior angles of parallelogram are supplementary ]

⇒ ${\displaystyle \frac{1}{2}}$ ∠DCB + ${\displaystyle \frac{1}{2}}$∠ABC = 90° 
⇒ ∠1 + ∠2 = 90° 
ΔCQB, ∠1 + ∠2 + ∠CQB = 180° 

From the above equation we get
∠CQB = 180° - 90° = 90° 
∠ RQP = 90°         ...[ ∠CQB = ∠ RQP , vertivally opposite angles ]
∠QRP = ∠RSP = ∠SPQ = 90° 
So, PQRS is a square.
Hence Proved.

Question 9

In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90^o.

Sol:

(i) Let AD = x
AB = 2AD = 2x
Also AP is the bisector ∠A
∠1 = ∠2
Now,
∠2 = ∠5                   ...[ alternate angles ]
Therefore ∠1 = ∠5
Now
AP = DP = x ...[ sides opposite to equal angles are also equal ]
Therefore
AB = CD ...[ opposite sides of  parallelogram are equal ]
CD = 2x
⇒ DP + PC = 2x
⇒ x + PC = 2x
⇒ PC = x
Also, BC = x
ΔBPC
⇒ ∠6 = ∠4 ...[ angles opposite to equal sides are equal ]
⇒ In ∠6 = ∠3
Therefore ∠3 =∠ 4
Hence BP bisect ∠B

(ii)
Opposite angles are supplementary
Therefore

∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ 2 ∠2 + 2 ∠3 =180°     .....[ ∠1 = ∠2 , ∠3 = ∠4 ]
⇒ ∠2 + ∠3 = 90°
ΔAPB
∠2 + ∠3 ∠APB = 180°
⇒ ∠APB = 180° - 90° ...[ by angle sum property ] 
⇒ ∠APB = 90° 
Hence proved.

Question 10

Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.

Sol:

Points M are N taken on the diagonal AC of a parallelogram ABCD such that.
Prove that BMDN is a parallelogram

construction: Join B to D to meet AC in O.

Proof: We know that the diagonals of a parallelogram bisect each other.
Now, AC and BD bisect each other at O.
OC = OA
AM = CN
OA - AM = OC - CN
OM = ON

Thus in a quadrilateral BMDN, diagonal BD and MN are such that OM = ON and OD = OB

Therefore the diagonals AC and PQ bisect each other.
Hence  BMDN is a parallelogram

Question 11

In the following figure, ABCD is a parallelogram.

Prove that:
(i) AP bisects angle A.
(ii) BP bisects angle B
(iii) ∠DAP + ∠BCP = ∠APB

Sol:

Consider ΔADP and ΔBCP,
AB = BC               ....[ Since ABCD is a parallelogram. ]
DC = AB               ....[ Since ABCD is a parallelogram. ]
∠A ≅ ∠C              ....[ Opposite angles ]
ΔADP ≅ ΔBCP     .....[ SAS ]

Therefore, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB,
AP = PB
∠APB = ∠DAP + ∠BCP
Hence proved

Question 12

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ;
prove that AP and DQ are perpendicular to each other.

Sol:

ABCD is a square and AP = PQ.

Consider ΔDAQ and ΔABP,
∠DAQ = ∠ABP = 90°
DQ = AP 
AD = AB
ΔDAQ ≅ ΔABP
⇒  ∠PAB = ∠QDA

Now,
∠PAB + ∠APB = 90°
also ∠QDA + ∠APB = 90°      ....[ ∠PAB = ∠QDA ]

Consider ΔAOQ by ASP
∠QDA + ∠APB + ∠AOD = 180° 
⇒ 90° + ∠AOD = 180° 
⇒ ∠AOD = 90° 
Hence AP and DQ are perpendicular.

Question 13

In a quadrilateral ABCD, AB = AD and CB = CD.
Prove that :
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Sol:

Given: ABCD is quadrilateral,
AB = AD
CB = CD

To prove:
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Proof:
In ΔABC and ΔADC,
AB = AD                    ....(given)
CB = CD                   .....(given)
AC = AC                   ......(Common side)
ΔABC ≅ ΔADC         .......(SSS)

∠BAD = ∠DAO       .......(AC bisects A)

Therefore AC bisects ∠BAD
OD = OB
OA = OC                 ......( diagonals bisect each other at O )
Thus AC is perpendicular bisector of BD.
Hence proved.

Question 14

The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC.

Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.

Sol:

Given ABCD is a trapezium, AB || DC and AD = BC.

Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.

Proof: (i) Since AD || CE and transversal AE cuts them at A and E respectively.
Therefore, ∠A + ∠B = 180°
Since, AB || CD and AD || BC
Therefore, ABCD is a parallelogram.
∠A = ∠C
∠B = ∠D              ....[ Since ABCD is a parallelogram ]
Therefore,
∠DAB = ∠CBA
∠ADC = ∠BCD

In ΔABC and ΔBAD, we have
BC = AD                 ....( given )
AB = BA                 ....( Common )
∠A = ∠B                 ....( proved )
ΔABC ≅ ΔBAD        ....( SAS )  
ΔABC ≅ ΔBAD
Since, Therefore AC = BD....( Corresponding parts of congruent triangles are equal. )
OA = OB
Again OC = OD      ....( Since diagonals bisect each other at O )
Hence proved.

Question 15

In the given figure, AP is the bisector of ∠A and CQ is the bisector of ∠C of parallelogram ABCD.

Prove that APCQ is a parallelogram.

Sol:

Construction: Join AC

Proof:
∠BAP = ${\displaystyle \frac{1}{2}}$∠A             ...( AP is the bisector  of ∠A )

∠DCQ = ${\displaystyle \frac{1}{2}}$∠C            ...( CQ is the bisector of ∠C ) 
⇒ ∠BAP = ∠DCQ           ....(i)....[ ∠A = ∠R ( Opposite angles of a parallelogram.) ]
Now,
∠BAC = ∠DCA                ....(ii)....[ Alternate angles since AB || DC ]

Subtracting (ii) from (i), We get
∠BAP - ∠BAC = ∠DCQ - ∠DCA 
⇒ ∠CAP = ∠ACQ
⇒ AP || QC                    .....( Alternate angles are equal )
Similarly, PC || AQ.
Hence, APCQ is a parallelogram.

Question 16

In case of a parallelogram
prove that:
(i) The bisectors of any two adjacent angles intersect at 90^o.
(ii) The bisectors of the opposite angles are parallel to each other.

Sol:

ABCD is a parallelogram, the bisectors of ∠ADC and ∠BCD meet at a point E and the bisectors of ∠BCD and ∠ABC meet at F.

We have to prove that the ∠CED = 90° and ∠CFG = 90°

Proof: In the parallelogram ABCD
∠ADC + ∠BCD = 180°       ....[ sum of adjacent angles of a parallelogram ]

⇒ ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°

⇒ ∠EDC + ∠ECD + ∠CED = 180°
⇒ ∠CED = 90°

Similarly taking triangle BCF it can be proved that ∠BFC = 90°
∠BFC + ∠CFG = 180°               ....[ adjacent angles on a line ]
Also ⇒ ∠CFG = 90°
Now since ∠CFG = ∠CED = 90° ....[ It means that the lines DE and BG are parallel ]
Hence proved.

Question 17

The diagonals of a rectangle intersect each other at right angles. Prove that the rectangle is a square.

Sol:

To prove: ABCD is a square,
that is, to prove that sides of the quadrilateral are equal
and each angle of the quadrilateral is 90°,
ABCD is a rectangle,
⇒ ∠A = ∠B = ∠c = ∠D = 90° and diagonals bisect each other.

that is, MD = BM                     ....(i)
Consider ΔAMD and ΔAMB,
MD =  BM                               ....( from(i) )
∠AMD = ∠AMB = 90°             .....(given)
AM = AM                                ......( common side )
ΔAMD ≅ ΔAMB                      ....(SAS ngruence iterion)
⇒ AD = AB                              ....( c.p.c.t.c. )
Since ABCD is a rectangle, AD = BC and AB = CD
Thus, AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D = 90°
⇒ ABCD is a square.

Question 18

In the following figure, ABCD and PQRS are two parallelograms such that ∠D = 120° and ∠Q = 70°.
Find the value of x.

Sol:

ABCD is a parallelogram.
⇒ Opposite angles of a parallelogram are congruent.
⇒ ∠DAB = ∠BCD and ∠ABC = ∠ADC = 120°
In ABCD,
∠DAB + ∠BCD + ∠ABC + ∠ADC = 360°   ....( sum of the measures of angles of a quadrilateral )
⇒ ∠BCD + ∠BCD + 120° + 120° = 360°
⇒  2∠BCD = 360° - 240°
⇒  2∠BCD = 120°
⇒  ∠BCD = 60°
PQRS is parallelogram.
⇒ ∠PQR = ∠PSR = 70°
In ΔCMS,
∠CMS + ∠CSM + ∠MCS = 180°      ....( angle sum property )
⇒ x + 70° + 60°  = 180° 
⇒ x  = 50°

Question 19

In the following figure, ABCD is a rhombus and DCFE is a square.

If ∠ABC =56°, find:
(i) ∠DAE
(ii) ∠FEA
(iii) ∠EAC
(iv) ∠AEC

Sol:

ABCD is a rhombus.
⇒ AD = CD and ∠ADC = ∠ABC = 56°
DCFE is a square.
⇒ ED = CD and ∠FED = ∠EDC = ∠DCF = ∠CFE = 90°
⇒ AD = CD = ED
In ΔADE,
AD = ED
⇒ ∠DAE = ∠AED                ...(i)
∠DAE + ∠AED + ∠ADE = 180°
⇒ 2∠DAE + 146° = 180°            ....( Since ∠ADE = ∠EDC + ∠ADC = 90° + 56° = 146° )
⇒ 2∠DAE = 34°
⇒ ∠DAE = 17°
⇒ ∠DEA = 17°                   ....(ii)

In ABCD,
∠ABC + ∠BCD + ∠ADC + ∠DAB = 360°
⇒ 56° + 56° + 2 ∠DAB = 360°   ....( ∵ Opposite angles of a rhombus are equal.)
⇒ 2∠DAB = 248°
⇒ ∠DAB = 124°
We know that diagonals of a rhombus, bisect its angles.

⇒ ∠DAC = ${\displaystyle \frac{124°}{2}}$ = 62°

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°
Now,
∠FEA = ∠FED - ∠DEA  
          = 90° - 17°             ....( From(ii) and each angle of a square is 90° )
        = 73°       
We know that diagonals of a square bisect its angles.
⇒ ∠CED = ${\displaystyle \frac{90°}{2}}$ = 45°
So,
∠AEC = ∠CED - ∠DEA
          = 45° - 17°
          = 28°
Hence, ∠DAE = 17°, ∠FEA = 73°, ∠EAC = 45° and ∠AEC = 28°.